Re: [NUMA , x86_64] Why memnode_shift is chosen with the lowest possible value ?

[Eric Dumazet <dada1@cosmosbay.com>]

Andi Kleen a écrit :
> On Friday 30 September 2005 11:09, Eric Dumazet wrote:
> 
>>+       while (populate_memnodemap(nodes, numnodes, shift + 1) >= 0)
>>+               shift++;
> 
> 
> 
> Why shift+1 here? 

Thank you Andi fo r reviewing this stuff

The idea it to find the highest shift value, and to break the loop as soon as 
the (shift + 1) value gives us an "shift too big" error.

Maybe you want to write :

         while (populate_memnodemap(nodes, numnodes, ++shift) >= 0) ;
	shift--;

Well, thats only style...


> 
> 
>>+               if ((end >> shift) >= NODEMAPSIZE)
>>+                       return 0;
> 
> 
> This should be >, not >= shouldn't it?

Let's take an example

end   = 0xffffffff;
start = 0xfff00000;
shift = 20
Suppose that NODEMAPSIZE == (end >> shift) == 0xfff

If the test is changed to :

if ((end >> shift) > NODEMAPSIZE)
	return 0;

We could do one of the iteration with (addr < end) but (addr >> shift) == 
NODEMAPSIZE

if (memnodemap[NODEMAPSIZE] != 0xff)
	return -1;
memnodemap[NODMAPSIZE] = i;

Thats bound violation of memnodemap[]

AFAIK, I wonder why NODEMAPSIZE is 0xfff and not 0x1000, because this off by 
one make half of memnodemap[] to be unused for power of two ram size.


> 
> -Andi
> 
> P.S.: Please cc x86-64 patches to discuss@x86-64.org

Ah thank you

Eric